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1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467.

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Presentation on theme: "1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467."— Presentation transcript:

1 1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore Mob:

2 2 Portal frames

3 3 Learning out Come Review of Design of Portal Frames Design example Continued

4 4 INTRODUCTION Step1: Design of slabs Step2: Preliminary design of beams and columns Step3: Analysis Step4: Design of beams Step5: Design of Columns Step6: Design of footings

5 5 PROBLEM 2

6 6 A portal frame hinged at base has following data: Spacing of portal frames = 4m Height of columns = 4m Distance between column centers = 10m Live load on roof = 1.5 kN/m2 RCC slab continuous over portal frames. Safe bearing capacity of soil=200 kN/m2 Adopt M-20 grade concrete and Fe-415 steel. Design the slab, portal frame and foundations and sketch the details of reinforcements.

7 7 Data given: Spacing of frames = 4m Span of portal frame = 10m Height of columns = 4m Live load on roof = 1.5 kN/m 2 Concrete: M20 grade Steel: Fe 415

8 8

9 9

10 10 Step1:Design of slab Assume over all depth of slab as 120mm and effective depth as 100mm Self weight of slab = 0.12 x 24 = 2.88 kN/m2 Weight of roof finish = 0.50 kN/m2 (assumed) Ceiling finish= 0.25 kN/m2 (assumed) Total dead load wd= 3.63 kN/m2 Live load wL= 1.50 kN/m2 (Given in the data) Maximum service load moment at interior support = = 8.5 kN-m

11 11 Step1:Design of slab (Contd) Mu lim =Q lim bd2 (Q lim =2.76) = 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > kN-m From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2 Spacing of 10 mm dia bars = (78.54 x 1000)/384= mm c/c Provide 200 c/c

12 12 Step1:Design of slab (Contd) Area of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm 2 Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c Provide 340 c/c. Main and dist. reinforcement in the slab is shown in Fig

13 13 Step1:Design of slab (Contd)

14 14 Step2: Preliminary design of beams and columns Beam: Effective span = 10m Effective depth based on deflection criteria = 10000/13 = mm Assume over all depth as 750 mm with effective depth = 700mm, breadth b = 450mm and column section equal to 450 mm x 600 mm.

15 15 Step3: Analysis Load on frame i) Load from slab = ( ) x 4 =20.52 kN/m ii) Self weight of rib of beam = 0.45x0.63x24= 6.80 kN/m Total  kN/m Height of beam above hinge = /2 =3.72 m The portal frame subjected to the udl considered for analysis is shown in Fig. 6.10

16 16 Step3: Analysis (Contd.)

17 17 Step3:Analysis (Contd) The moments in the portal frame hinged at the base and loaded as shown in Fig. is analised by moment distribution IAB = 450 x /12 = 81 x 10 8 mm4, IBC= 450 x /12 = x 10 8 mm4 Stiffness Factor: KBA= IAB / LAB = x 10 5 KBC= IBC / LBC = 15.8 x 10 5

18 18 Step3:Analysis (Contd) Distribution Factors: Fixed End Moments: MFAB= MFBA= MFCD= MFDC 0 MFBC= -=-233 kN-m and MFCB= =233 kN-m

19 19 Step3:Analysis (Contd) Moment Distribution Table

20 20 Step3:Analysis (Contd) Bending Moment diagram

21 21 Step3:Analysis (Contd) Design moments: Service load end moments: M B =156 kN-m, Design end moments M uB =1.5 x 156 = 234 kN-m, Service load mid span moment in beam= 28x10 2 /8 – 102 =194 kN-m Design mid span moment Mu+=1.5 x 194 = 291 kN-m Maximum Working shear force (at B or C) in beam = 0.5 x 28 x 10 = 140kN Design shear force Vu = 1.5 x 140 = 210 kN

22 22 Step4:Design of beams: The beam of an intermediate portal frame is designed. The mid span section of this beam is designed as a T-beam and the beam section at the ends are designed as rectangular section. Design of T-section for Mid Span : Design moment Mu=291 kN-m Flange width bf= Here Lo=0.7 x L = 0.7 x 10 =7m bf= 7/ x0.12=2.33m

23 23 Step4:Design of T-beam: b f /b w =5.2 and D f /d =0.17 Referring to table 58 of SP16, the moment resistance factor is given by K T =0.43, M ulim =K T bwd 2 fck = 0.43 x 450 x x 20/1x10 6 = kN-m > Mu Safe The reinforcement is computed using table 2 of SP16

24 24 Step4:Design of T- beam: Mu/bd 2 = 291 x 10 6 /(450x700 2 )  1.3 for this p t =0.392 A st =0.392 x 450x700/100 = mm 2 No of 20 mm dia bar = /(  x20 2 /4) =3.93 Hence 4 Nos. of #20 at bottom in the mid span

25 25 Step4:Design of Rectangular beam: Design moment M uB =234 kN-m M uB /bd 2 = 234x106/450x700 2  1.1 From table 2 of SP16 p t =0.327 A st =0.327 x 450 x 700 / 100 = 1030 No of 20 mm dia bar = 1030/(  x20 2 /4) =3.2 Hence 4 Nos. of #20 at the top near the ends for a distance of o.25 L = 2.5m from face of the column as shown in Fig

26 26 Step4:Design of beams Long. Section:

27 27 Step4:Design of beams Cross-Section:

28 28 Check for Shear: Nominal shear stress = pt=100x 1256/(450x700)=0.39  0.4 Permissible stress for pt=0.4 from table 19  c=0.432 <  v Hence shear reinforcement is required to be designed Strength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kN Shear to be carried by steel Vus= = 74 kN

29 29 Check for Shear: Nominal shear stress = p t =100x 942/(400x600)=0.39  0.4 Permissible stress for pt=0.4 from table 19  c =0.432 <  v Hence shear reinforcement is required to be designed Strength of concrete V uc =0.432 x 400 x 600/1000 = 103 kN Shear to be carried by steel V us = = 59 kN

30 30 Check for Shear: Spacing 2 legged 8 mm dia stirrup sv= Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

31 31 Step5:Design of Columns: Cross-section of column = 450 mm x 600 mm Ultimate axial load Pu=1.5 x 140 = 210 kN (Axial load = shear force in beam) Ultimate moment Mu= 1.5 x 156 = 234 kN-m ( Maximum) Assuming effective cover d’ = 50 mm; d’/D  0.1

32 32

33 33 Step5:Design of Columns: Referring to chart 32 of SP16, p/fck=0.04; p=20 x 0.04 = 0.8 % Equal to Minimum percentage stipulated by IS (0.8 % ) Ast=0.8x450x600/100 = 2160 mm2 No. of bars required = 2160/314 = 6.8 Provide 8 bars of #20

34 34 Step5:Design of Columns: 8mm diameter tie shall have pitch least of the following Least lateral dimension = 450 mm 16 times diameter of main bar = 320 mm 48 times diameter of tie bar = mm Provide 8 mm 300 mm c/c

35 Tie c/c 8-#20 450

36 36 Step6:Design of Hinges: At the hinge portion, concrete is under triaxial stress and can withstand higher permissible stress. Permissible compressive stress in concrete at hinge= 2x0.4f ck =16 MPa Factored thrust =P u =210kN Cross sectional area of hinge required = 210x10 3 /16=13125 mm 2 Provide concrete area of 200 x100 (Area =20000mm 2 ) for the hinge

37 37 Step6:Design of Hinges: Shear force at hinge = Total moment in column/height = 156/3.72=42 Ultimate shear force = 1.5x42=63 kN Inclination of bar with vertical =  = tan -1 (30/50) =31 o Ultimate shear force = 0.87 f y A st sin  Provide 4-#16 (Area=804 mm 2 )

38 38 Step7:Design of Footings: Load: Axial Working load on column = 140 kN Self weight of column =0.45 x 0.6 x3.72x 24 = 24 Self weight of = 16 kN Total load = 180 kN Working moment at base = 42 x 1 =42 kN-m

39 39 Step6:Design of Footings: Approximate area footing required = Load on column/SBC = 180/200 =0.9 m 2 However the area provided shall be more than required to take care of effect of moment. The footing size shall be assumed to be 1mx2m (Area=2 m 2 )

40 40 2mX 1m 0.7m 0.6m X 0.45m

41 41 Step6:Design of Footings: Maximum pressure qmax=P/A+M/Z = 180/2+6x42/1x22 = 153 kN/m 2 Minimum pressure qmin =P/A-M/Z = 180/2-6x42/1x2 2 = 27 kN/m 2 Average pressure q = (153+27)/2 = 90 kN/m 2 Bending moment at X-X = 90 x 1 x 0.72/2 = 22 kN-m Factored moment Mu  33 kN-m

42 42 Step6:Design of Footings: Over all depth shall be assumed as 300 mm and effective depth as 250 mm, Corresponding percentage of steel from Table 2 of SP16 is pt= 0.15% > Minimum p t =0.12%

43 43 Step6:Design of Footings: Area of steel per meter width of footing is A st =0.12x1000x250/100=300 mm2 Spacing of 12 mm diameter bar = 113x1000/300 = 376 mm c/c Provide 300 c/c both ways

44 44 Step6:Design of Footings: Length of punching influence plane = a o = = 850 mm Width of punching influence plane = b o = = 700 mm Punching shear Force = V punch =180-90x(0.85x0.7)=126.5 kN Punching shear stress  punch =V punch /(2x(a o +b o )d) =126.5x103/(2x( )250) = 0.16 MPa Permissible shear stress = 0.25  fck=1.18 MPa >  punch Safe

45 45 Step6:Design of Footings: Check for One Way Shear Shear force at a distance ‘d’ from face of column V= 90x1x0.45 = 40.5 kN Shear stress  v =40.5x103/(1000x250)=0.162 MPa For p t =0.15, the permissible stress  c = 0.28 (From table 19 of IS ) Details of reinforcement provided in footing is shown in Fig.

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48 48 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore Mob:


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